Problem: In Takahashi's mind, there is always an integer sequence of length 2 \times 10^9 + 1: A = (A_{-10^9}, A_{-10^9 + 1}, ..., A_{10^9 - 1}, A_{10^9}) and an integer P.

Initially, all the elements in the sequence A in Takahashi's mind are 0, and the value of the integer P is 0.

When Takahashi eats symbols `+`, `-`, `>` and `<`, the sequence A and the integer P will change as follows:

* When he eats `+`, the value of A_P increases by 1;
* When he eats `-`, the value of A_P decreases by 1;
* When he eats `>`, the value of P increases by 1;
* When he eats `<`, the value of P decreases by 1.



Takahashi has a string S of length N. Each character in S is one of the symbols `+`, `-`, `>` and `<`. He chose a pair of integers (i, j) such that 1 \leq i \leq j \leq N and ate the symbols that are the i-th, (i+1)-th, ..., j-th characters in S, in this order. We heard that, after he finished eating, the sequence A became the same as if he had eaten all the symbols in S from first to last. How many such possible pairs (i, j) are there?

Constraints

* 1 \leq N \leq 250000
* |S| = N
* Each character in S is `+`, `-`, `>` or `<`.

Input

Input is given from Standard Input in the following format:


N
S


Output

Print the answer.

Examples

Input

5
+>+<-


Output

3


Input

5
+>+-<


Output

5


Input

48
-+><<><><><>>>+-<<>->>><<><<-+<>><+<<>+><-+->><<


Output

475

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define ull unsigned long long
#define key 1000000007
#define k 9223372036854775807ull
using namespace std;
int n;
char s[250010];
map <ull, ull> M;
inline ull power(ull x, ull y)
{
	ull Ans = 1;
	while(y)
	{
		if(y & 1)
			Ans *= x;
		y >>= 1;
		x *= x;
	}
	return Ans;
}
map <ull, ull> P;
int main()
{
	ull ny = power(key, k);
	scanf("%d%s", &n, s);
	ull w = 0;
	ull p = 0;
	for(int i = 0; i < n; ++ i)
	{
		if(s[i] == '+')
			++ P[p];
		else if(s[i] == '-')
			-- P[p];
		else if(s[i] == '>')
			++ p;
		else -- p;
	}
	for(int i = - n; i < 0; ++ i)
		w += power(ny, - i) * P[i];
	for(int i = 0; i <= n; ++ i)
		w += power(key, i) * P[i];
	ull a = 1, b = 0;
	ull Ans = 0;
	for(int i = n - 1; i >= 0; -- i)
	{
		++ M[-b * power(a, k)];
		if(s[i] == '+')
			++ b;
		else if(s[i] == '-')
			-- b;
		else if(s[i] == '>')
			a *= key, b *= key;
		else if(s[i] == '<')
			a *= ny, b *= ny;
		Ans += M[(w - b) * power(a, k)];
	}
	printf("%llu", Ans);
	return 0;
}